I would be very grateful, if someone could explain where and why I am wrong.

Just to repeat the question:Assets A and B are worth 100 today. Asset A will be worth 110 tomorrow with probability 0.9 and 90 otherwise. Asset B will be worth 110 with probability 0.5 and 90 otherwise. Asset C is worth 1 both today and tomorrow.How will the price of a call options on A and B struck at 100 compare.

My first minor question is what Asset C has got to do with the Asset A and B? Why is is introduced into the question?

Now my incorrect solution (only for Asset A, for Asset B same process just different numbers):

Value of our Portfolio in world A: 0.9*(d*110-10)

Value in world B: 0.1*d*90

To hedge the risk the value of the portfolio should be the same in either state of the world:0.9*(d*110-10) = 0.1*d*90d = 0.1 (d = number of shares)

I never did understand why Asset C was mentioned at all in this problem (maybe Prof Joshi could explain this for me as well?). The only thing I could think of is that asset C is implicitly assumed to be our choice of numeraire? (When doing risk-neutral pricing we do so relative to some numeraire; otherwise the presence of asset C makes no sense to me).

However, your approach to this problem is wrong, or at least it's the wrong way to think about it, because it assumes that the real world probabilities actually matter. The miracle here is that they don't! The whole idea behind option pricing is that we can use the risk-neutral probability measure and get the correct price; we don't need to use the real world probabilities at all.

So all the effort you're putting into this to solve the problem that way won't work. The way to do this is to change the probabilities in both option A and option B to risk neutral probabilities.

The risk-neutral probability is found by:

pS_{+} + (1-p)S_{-} = S_0

What that means is that in the risk-neutral measure, the expected value of the stock price tomorrow is just the stock price today (here S_{+} is the value if we experience an up-move, and S_{-} is the value of a down-move).

So in each case (case A and case B), the value of S_{+}=110 and the value of S_{-}=90, and the value of S_0=100. So p will be the same in each case (although it doesn't matter in this instance what p is, you can see that p=1/2). Since the probabilities will be the same, the price (which is the risk-neutral expected value of the payoff) will also be the same.

The point of this exercise is for you to recognize that if we have the same basic properties of the assets, and when the only thing that differs is the probability of up or down moves, then in the risk neutral measure they have the same properties, and hence options will give the same price. (Both will be worth 5; just take the expected value).

I would recommend you think a long time on this; I'm now in chapter 8 of this book and this is a concept that will not go away (in fact you'll see in chapter 6 that Mr Joshi himself calls this risk-neutral pricing formula "the most important equation in mathematical finance").

Thank you very much for all the detailed answers. The help is much appreciated. I have to admit that I still find it counterintuitive to use risk-neutral probabilities and not real-world probabilities

Let's assume the probability for Asset A reaching 110 is 99.999% and I was offered an option for 5 to buy it for a price of 100. This would seem a very good deal for me. However, the seller then would have to sell at a lower price than the asset is worth.

Turning it around, now the probability is 99.999% for a value of 90. In this case the option is nearly "worthless", as it wouldn't be exercised.

Why not use real world probabilities, as they determine the likelihood that an asset reaches certain price. If I use p=0.5 I can see that both State A and B are equally likely, but that is a "fictitious world" and not the real world?

The way this works is that in a one-step, two-state model, you can hedge. Prof. Joshi does this a bit in his book, but you see a bunch of this in Shreve's Volume I (which I'll be honest I never bothered to read). Here's how you might think about it. Note: We are given asset C as the riskless asset, and it's worth 1 both today and tomorrow so you can think of this as a case where interest rates are 0.

The way to approach this, in a general setting, is to take some initial wealth, determine how many shares of stock you need to buy to hedge the risk, and take the remaining amount and invest in the money market. Let X(0) be how much you have initially, g be the number of shares (since I don't have a delta on my keyboard; sorry, you'll just have to deal), S(0) the value of asset A at time 0, and S(1) the value of asset A at time 1.

Then our portfolio at time 1 is worth:

gS(1) + (X(0)-gS(0)) = X(0) + g(S(1)-S(0))

We want to choose X(0) and g so that X(1) models the payoff of our option.

Now to purchase a unit of asset A costs $100. As you see above, we need g=1/2, which would cost $50. We also saw that X(0) = 5; then it costs $50 to buy half a share of asset A and we would need to short 45 units of asset C. So what are your possible values of our portfolio at time 1?

In the up case, the value of asset A will be 110, so 1/2 a share will be worth 55. In the down case, half a share will be worth 45. So X(1) now models perfectly the payoff of this call option struck at 100 (it pays 10 on an up-move and pays 0 on a down-move). This suggests an option price of 5 should work.

So now what? We need to introduce probabilities into the picture. However, the reason why your real-world probability choice is irrelevant is because of this ability to hedge. The fact that we can hedge does, in effect, remove the risk associated with those probabilities (that's kind of what hedging means ).

Let us now take X(0) our initial wealth and g the number of shares we purchase. We want to determine X(0) and g needed to perfectly model this system, using probabilities. In an up-move, X(1) = X(0) + (110-100)g = 10 and in a down-move, X(1) = X(0) + (90-100)g = 0

If we take two numbers p and q with p+q=1, then we can actually take:

X(0) + g[p(110) + q(90) - 100] = p(10)-q(0)

(we find this by just multiplying X(1) by p in an up-move case and by q in a down-move case and add them up). If we choose p so that p(110) + q(90) - 100 = 0, then we have chosen p so that 5 = 10p, which shows that p=1/2 afterall.

In general, suppose you have a set amount of wealth X(0); then you can do a similar trick by determining how many shares of your asset you would need to purchase in order to hedge your risk. After doing so, you should be able to replicate your option by constructing a portfolio that matches its payoff, and then use the above trick to determine how much the option would cost by determining the expected value of X(1) which can be done using those values of p and q you determine above.

So to summarize, this is indeed fictitious, but in some sense it actually isn't, because the ability to perfectly hedge your risk allows you to remove that risk. That's why you can use risk-neutral pricing. Note that this is why Prof. Joshi said that you needed asset C; without a riskless asset to work with, you can't hedge your risk and therefore you can really talk about risk-neutral pricing.

Anyways, I think that more or less is what is going on. Sorry for answering questions on here, but I'm using this as a chance to test my own understanding also. Maybe Prof Joshi could tell me if I'm more or less right here?

Seems to me if the stock will almost surely drop, there's almost no value in a call option struck at today's price. If it will rise almost surelly to x then the price of the call option should be almost x-100.

I guess that's why it doesn't make sense to value it at 5$ in these cases. Thing is, in risk-neutral pricing, the correct option call price matches the risk-free return. But if p is very close to 0 or 1 then everything is risk free again. But in a risk-neutral world options don't make sense as an effective concept I guess.

If the call option is struck at zero then the call option price should be today's price.

I'm not sure I got your last statement, which part of the argument are you referring to? Still getting used to some of these concepts, guess I missed smthg along the way.

Sure I agree that the stock prices become as unreasonable as the option prices, in the p~0 and p~1 scenarios.I also seems clear that the call option price, when the struck price is 0, should be the just the stock price due to shorting.

Well if one is more reasonable than the other there should be some sort of arbitrage, no?

By the way, if I'm selling to you a call option struck at zero, and valuing it at the stock price as I should, then isn't it true that I'm just short selling? I get the call option price from you today which would be 100$ and you will surely call it since the struck price is zero so I'll have to give you the stock tomorrow.